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skad [1K]
3 years ago
9

A bird can fly at an elevation of 6400 meters. A fish can swim at a depth of 500 meters. What is the difference between the bird

s elevation and the fish’s depth
Mathematics
2 answers:
never [62]3 years ago
5 0

The depth can be represented as

d = b - f where

d = depth

b = bird height

f = fish depth

d = 6400 - (- 500)

d = 6400 + 500

d = 6900 m

Answer: <em><u>The difference in the elevation and depth of the bird and fish is 6900 m</u></em>

Inessa [10]3 years ago
4 0

5900 is  the answer because

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A coin is pulled from a jar and then put back in the jar. If 40 coins are pulled, enter the number of times a nickel is expected
neonofarm [45]

Answer:

20 times

Step-by-step explanation:

40 divided by 2

4 0
3 years ago
2 3/5+2 4/5 I need the answer
gavmur [86]

Answer:

5 2/5

Step-by-step explanation:

2 3/5 + 2 4/5

= 13/5 + 14/5

= 27/5

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5 0
3 years ago
Read 2 more answers
Write an equation of the line that passes through the point (1,3) and has a slope of -2
jek_recluse [69]

Answer:

y= -2x+1

Step-by-step explanation:

y=mx+b

y= -2x +b

3= -2(1) +b

3= -2 +b

b=1

y= -2x+1

4 0
2 years ago
Help please with algebra problem
Strike441 [17]
Number of weekend minutes used: x
Number of weekday minutes used: y

This month Nick was billed for 643 minutes:
(1) x+y=643

The charge for these minutes was $35.44
Telephone company charges $0.04 per minute for weekend calls (x)
and $0.08 per minute for calls made on weekdays (y)
(2) 0.04x+0.08y=35.44

We have a system of 2 equations and 2 unkowns:
(1) x+y=643
(2) 0.04x+0.08y=35.44

Using the method of substitution
Isolating x from the first equation:
(1) x+y-y=643-y
(3) x=643-y

Replacing x by 643-y in the second equation
(2) 0.04x+0.08y=35.44
0.04(643-y)+0.08y=35.44
25.72-0.04y+0.08y=35.44
0.04y+25.72=35.44

Solving for y:
0.04y+25.72-25.72=35.44-25.72
0.04y=9.72

Dividing both sides of the equation by 0.04:
0.04y/0.04=9.72/0.04
y=243

Replacing y by 243 in the equation (3)
(3) x=643-y
x=643-243
x=400


Answers:
The number of weekends minutes used was 400
The number of weekdays minutes used was 243
6 0
3 years ago
At the start of the year, 15 chameleons were introduced into a zoo. The population of chameleons is expected to grow at a rate o
bearhunter [10]

Answer:

option-B

Step-by-step explanation:

We are given

At the start of the year, 15 chameleons were introduced into a zoo

so, P_0=15

The population of chameleons is expected to grow at a rate of 41.42% every year

so, r=0.4142

and x represents the number of years since the chameleons were introduced into the zoo

now, we can set equation to find total population

and we get

P(x)=P_0(1+r)^x

now, we can plug values

P(x)=15(1+0.4142)^x

P(x)=15(1.4142)^x

Average rate of change between 2 years and 4 years:

we can use formula

A_1=\frac{P(4)-P(2)}{4-2}

now, we can plug values

A_1=\frac{15(1.4142)^{4}-15(1.4142)^{2}}{4-2}

A_1=14.99914

Average rate of change between 4 years and 6 years:

we can use formula

A_2=\frac{P(6)-P(4)}{6-4}

now, we can plug values

A_2=\frac{15(1.4142)^{6}-15(1.4142)^{4}}{6-4}

A_2=29.99770

Average rate of change between 6 years and 8 years:

we can use formula

A_3=\frac{P(8)-P(6)}{8-6}

now, we can plug values

A_3=\frac{15(1.4142)^{8}-15(1.4142)^{6}}{8-6}

A_3=59.99425

now, we will check each options

option-A:

we can see that

A_3-A_2=30

A_3-A_2=30

So, this is FALSE

option-B:

A_1=\frac{1}{2}A_2

So, this is TRUE

option-C:

This is FALSE

option-D:

we got

A_1=\frac{1}{2}A_2

so, this is FALSE


5 0
3 years ago
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